RE: Error?
- To: mathgroup at smc.vnet.net
- Subject: [mg44717] RE: [mg44680] Error?
- From: "Florian Jaccard" <florian.jaccard at eiaj.ch>
- Date: Sat, 22 Nov 2003 02:17:12 -0500 (EST)
- Reply-to: <florian.jaccard at eiaj.ch>
- Sender: owner-wri-mathgroup at wolfram.com
Hello !
Very interesting question !
As
Integrate[Sqrt[r^2 - x^2], x] = (1/2)*(x*Sqrt[r^2 - x^2] +
r^2*ArcTan[x/Sqrt[r^2 - x^2]])
you will have to use complex numbers using Newton-Leibnitz's formula with
R<d<2R !
As (I suppose) you want to compute a surface under a circle, it is not good
!
so I would suggest that you should inetgrate only with d<R !
Indeed :
In[34]:=
Integrate[Sqrt[r^2 - x^2], {x, -r, -r + d}, Assumptions -> {d > 0, r > 0, d
< r}]
Out[34]=
(1/4)*(2*Sqrt[(-d)*(d - 2*r)]*(d - r) + Pi*r^2 + 2*r^2*ArcSin[(d - r)/r])
So if d>R, just compute for x from 0 to d-R and add Pi r^2/4 !
Meilleures salutations
Florian Jaccard
-----Message d'origine-----
De : Baris Altunkaynak [mailto:altunkai at boun.edu.tr]
Envoyé : ven., 21. novembre 2003 11:13
À : mathgroup at smc.vnet.net
Objet : [mg44680] Error?
This integral below give 0 on Mathematica 5.0
Integrate[Sqrt[R^2 - x^2], {x, -R, -R + d},
Assumptions -> R > 0 && d > 0 && d < 2R]
This is a mistake I think, isn't it?
Thanks,
Baris Altunkaynak